NY Regents Chemistry Exam June 2023 Questions 70-73

The Ostwald process is an industrial method to produce nitric acid, HNO3(aq), used in the manufacture of fertilizers. Several steps are involved in this process. In the first step, ammonia and oxygen react in the presence of a catalyst, as represented by unbalanced equation 1.

In the second step, nitrogen(II) oxide reacts with oxygen to produce nitrogen(IV) oxide,

represented by balanced equation 2 below.

Equation 2: 2NO(g) + O2(g) → 2NO2(g) + heat

70 Determine the percent composition by mass of nitrogen in HNO3 (gram-formula mass= 63.0 g/mol).

*Solution: The formula for calculating percent composition by mass can be found in Table T on the Reference Table. It is % composition by mass = (mass of part)/(mass of whole) 100%. Mass of whole is given (63.0 g/mol). Mass of nitrogen can be found on the Periodic Table. We get (14/63)100% = 22.2%*

71 Balance equation 1 *in your answer booklet*, using the smallest whole-number coefficients.

*Solution:NH3(g) + O2(g) → NO(g) + H2O(g) + heat is an unbalanced equation.*

*We balance equations by putting coefficients in front of molecules to make sure there is an equal number of atoms of each element on both sides. There are 3 hydrogens on the left and 2 on the right. To balance we put 2 in front of NH3 and 3 in front of H2O and get*

*2NH3(g) + O2(g) → NO(g) + 3H2O(g) + heat*

*Now the hydrogen is balanced but the nitrogen is not. We need to put 2 in front of NO to balance N.*

*2NH3(g) + O2(g) → 2NO(g) + 3H2O(g) + heat*

*Lastly, there are 2 oxygens on the left and 5 on the left. We need to use a fraction (5/2) to make them equal.*

*2NH3(g) + (5/2)O2(g) → 2NO(g) + 3H2O(g) + heat*

*To get rid of the fraction, we need to multiply everything by 2 and get*

*4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) + heat*

72 Show a numerical setup for calculating the gram-formula mass of the NO2(g) produced in equation 2.

*Solution: To calculate the gram-formula mass we multiply the number of atoms of each element by the mass taken from the Periodic Table.*

*NO2*

*N=1*14.0 g/mol*

*O = 2*16.0 g/mol*

*Gram formula mass of NO2 = 1*14.0 g/mol+ 2*16.0 g/mol*

73 Determine the number of moles of oxygen required to completely react with 4.0 moles of NO(g) in equation 2.

*Solution: We can use the coefficients in equation 2 to represent moles. There are 2 moles of NO needed to react with 1 mol of O2. If we need to use 4 moles of NO, 2 moles of O2 will be needed. *

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